Le Fevre High School
SACE Stage 2 Physics
Motion of Charged Particles in Magnetic Fields Solutions
1.
A proton moves vertically down at 6.7 x 107 m s-1 through a north-south
magnetic field of 7.3 x 10-3 T. Calculate the magnitude and direction of the force
that acts on the proton.
N
v down
B
F
W
E
F = B q v sin
= 7.3 x 10 -3 x 1.6 x 10 -19 x 6.7 x 10 7 x sin90o
F = 7.83 x 10 -14 N
West
S
2.
A positively charged carbon ion (mass 2.00 x 10-26 kg and charge 1.60 x 10-19 C)
moves in a circular path of radius 0.20 m. A magnetic field of 0.50 T is
perpendicular to the plane of the circular path. Calculate the speed of the
carbon ion.
F = Fmagnetic
mv 2
=Bqv
r
Bqr
 v=
m
0.05 x1.6 x1019 x 0.20
v =
2.0 x10 26
= 8.0 x 10 5 m s -1
3.
A certain charge moving north at 3.00 x 102 m s-1 enters a region in which there
is a 5.00 x 10-6 T field acting vertically downward. The charge undergoes
circular motion in a clockwise direction with a centripetal force of 3.00 x 10-3 N
acting. Find the nature and size of the charge.
F = Fmagnetic
3 x 10 -3 = B q v
3 x 10 3
q=
Bv
F
3 x 10 3
q=
5 x 10 6 x 3 x10 2
v
q = -2 C
Use right hand rule to determine that sign is
NEGATIVE .
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4. A proton is accelerated from rest through a potential difference of 1.7 x 105 V and
enters a magnetic field having a strength of 0.20 T. Determine the:
(a) speed of the proton,
initial K of proton = 170000 eV (proton has same magnitude charge as an
electron). Hence loss in electric potential Energy = gain in K
1
170000 x 1.6 x 10 -19 = x 1.67 x 10-27 x v 2
2
-14
2.72 x 10 = 8.35 x 10-28 x v2
v2 = (2.72 x 10-14) / (8.35 x 10-28)
v = 5.71 x 10 6 m s -1
(make sure you use the mass of a proton not an electron, but note that the charge on a
proton is the same size as the charge on an electron but is of course positive)
(b) radius of the path of the proton,
F = Fmagnetic
mv 2
=Bqv
r
mv
r=
Bq
1.67 x 10 27 x 5.71 x 10 6
r=
0.20 x 1.6 x 10 19
r = 0.298  0.30 m
(c) time for one revolution of the proton.
v=
2r
T
2r circumfere nce
=
v
v
2 x 0.298
T=
5.71 x 10 6
= 3.3 x 10 -7 s
T=
5.
An electron moving at 7.00 x 105 m s-1 enters a region in which a uniform
magnetic field exists and describes a circle of radius of 0.640 m. Find:
(a) the magnetic field strength in this region,
(b) the period of revolution of the electron.
6.
As seen in the diagram over the page, a beam of particles of charge q enters a
region where an electric field is uniform and directed downward. Its value is 80
kV m-1.
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(a) Write down an expression for the magnitude of the force ‘F’ acting on the
charge due to the electric field in terms of ‘q’ and the electric field strength
given.
F = Eq
E = 80 kV/m = 8.0 x 104 V m-1 down
F = 8.0 x 104q towards the top of the page.
Perpendicular to E and directed into the page is a magnetic field B = 0.4 T.
(b) Write down an expression for the magnetic force acting on the charge in terms
of ‘q’, the speed ‘v’ and the magnetic field strength given.
F = Bqvsin = Bqv = 0.4qv T ( = 90o)
B = 0.4 T into page
If the speed of the particles is properly chosen, the force due to the electric field
and that due to the magnetic field will be equal in magnitude. What speed is
selected in this case? (This device is called a velocity selector.)
for undeflected motion
Fmagnetic = Fcoulombic
0.4qv = 8.0 x 104q
v=
8.0 x 10 4
0 .4
v = 2.0 x 105 m s-1
7.
A beam of cathode rays (electrons) is found to move in an arc of radius 0.45 m
in a magnetic field of strength 300 T. Calculate the speed of the cathode rays.
r = 0.45 m
B = 300 T = 300 x 10-6 T
q = 1.6 x 10-19 C
mv 2
= qvB
r
BqR
m
(3.00 x 10 4 )(1.6 x 10 19 )(0.45)
v=
9.11 x 10 31
 v=
v = 2.37 x 107 m s-1
from magnetic field deflection,
Fcent = Fmag
mv 2
= qvB
r
8.
The A cyclotron has a radius r and a magnetic field of strength B. It is used to
accelerate particles of mass m, carrying a charge q.
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(a)
Show that the period T of the motion of the particles in their circular
2m
paths is given by T 
.
qB
Radius of the circular path in the magnetic field is,
mv
r
qB
rqB
v 
m
The time to traverse a semicircle,
r
t
v
rm
t 
rBq
m
t 
Bq
Therefore, for a complete circle, x2 
2m
t
Bq
(b)
Show that the kinetic energy K of the emerging particles is given by
q2 B2r 2
K
2m
EK 
1
2
mv 2
 rBq 
 E K  2 m

 m 
1 r 2 B2q2
 EK  2
m
2 2 2
r B q
 EK 
2m
1
2
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9.
the
The Dees of a cyclotron have a diameter of 90cm. The magnetic field inside
evacuated Dees is 1.9T. This cyclotron is used to accelerate protons.
(a)
Find the period of the motion of the protons.
2m
T
Bq
2  1.673  10  27
1.9  1.6  10 19
 3.45  10 8 s

(b)
be
What is the frequency of the alternating potential difference that must
applied to the Dees?
1
1
f  
T 3.45  10 8
 3  10 7 Hz
Therefore f = 6 x 107Hz (jumps gap twice per
revolution)
(c)
from
Determine the energy (in electron volts) of the protons that emerge
this cyclotron.
q2 B2r 2
EK 
2m
(1.6  10 19 ) 2  (1.9) 2  (0.45) 2

2  1.673  10  27
 5.6  10 12 J
 3.5  10 7 eV
 35MeV
(d)
Consider a cyclotron that is identical to the one above except that its
diameter is twice as great. What effect would this have on,
(i)
the period of the protons.
2m
T
Bq
ie, T does not depend on the diameter therefore no effect!
(ii)
the kinetic energy of the protons.
q2 B 2r 2
EK 
2m
2
EK  r
double r, 4 times EK
Therefore K is 4 times bigger.